[SOLVED] Com s 228 programming project 5 generating a perfect hash table implementation

Graphs are the most general of the common data structures; consider that a scalar object is a degenerate list,
which is a degenerate tree, which is itself a degenerate graph, or in other words, every data structure we’ve
discussed this semester can be represented as a graph. Due to their extreme generality, it’s rare that you’ll
use a generic “graph” data structure.The limited set of things you might do with a list make two generic
list implementations—an array based- and a linked-structure—good enough for most implementations, and
it’s only if you are doing, e.g., low-level system programming that you might have a good reason to write a
custom, one-off list. But there is so much variation in graphs that it is perhaps impossible to write a good
and useful generic graph implementation.Hash tables provide a means to get constant-time look-up performance on a collection. Usually, the
data comes to us at runtime, often in an on-line fashion. In this case, we take what foreknowledge we can
to inform best practices in hash table construction, choosing a hash function, etc., to build a data structure
with as close to optimal performance as possible; however, if the keys in the table are known beforehand,
it’s always possible to construct a perfect hash table, a table in which every key maps to a unique slot.There are many algorithms to construct such a table and the associated hash function; almost all of them are
implemented using graphs.For this project, you will be implementing a perfect hash table construction algorithm1
. This algorithm
generates a pair of tables of random numbers. Keys are hashed with the values in each these tables, giving
two integers (one for each table).These integers are then taken as node indices in a graph, with an edge between them. So long as the resultant graph—after hashing all of the keys—contains no cycles, the algorithm
is ready to generate the hash table as a function of the tables (the T tables) used in the graph generation and
a function g() on those tables.Algorithm Example
NOTE: This algorithm looks pretty intimidating at first glance. With some careful reading and workingthrough of examples, you should find it’s actually fairly straightforward. That said, you don’t necessarily
have to understand it at all! If you find yourself getting anxious about all this “scary-looking stuff”, jump
down to the Requirements sections, read that, then come back here with a more relaxed mindset. It is very
common for professional programmers to be required to implement solutions to problems that they don’t
actually understand, so if you find yourself doing that, here or elsewhere, please know that it is normal.This
is not to advocate for ignorance, however–it’s always better to understand something than not to understand
it–only to recognize that sometimes the cost-benefit of understanding is too high!The example below was generated by hand, and uses an optimization that would be very difficult to
implement in program control. We note that our original keys (the number names from “one” to “ten”)
1The algorithm is presented in: Zbigniew J. Czech, George Havas, and Bohdan S. Majewski, “An optimal algorithm for generating minimal perfect has functions”, Information Processing Letters, 43(5):257-264, October 1992.are unambiguously differentiable by the first two letters. Thus, we can simplify the presentation by using
only the first two letters of the keys. In your Java program, you will always use the entirety of every word;
however, the method that hashes the words is already implemented for you, so you don’t need to concern
yourself with it unless you are endeavoring to fully understand the implementation.Our keys and the “sub-keys” that we’re using in this example follow:
Key Minimum Unique Sub-key
one on
two tw
three th
four fo
five fi
six si
seven se
eight ei
nine ni
ten teA modulus must be chosen, giving a maximum number in our tables, and a maximum number of nodes
in the resultant graph. The paper authors choose a value that is one more than twice the number of keys,
so we do too. This choice has implications on the runtime of the algorithm (smaller moduli increase the
probability of cycles in the resultant graph, forcing a restart of the algorithm). If the modulus is too small,
the algorithm will enter an infinite loop.Next, we populate our tables, T1 and T2, with random numbers less than the modulus. Each of the two
tables will have the same number of rows as the length l of the sub-key. The i
th row, 0 <= i <= l − 1,
defines a separate mapping from letters at the i
th position (increasing from left to right) within the sub-key
to random numbers.T1
e f h i n o s t w
8 15 10 2 10 19 13 0 13
3 5 0 5 5 6 19 5 6
T2
e f h i n o s t w
20 2 4 19 11 15 3 8 11
16 4 2 6 19 10 18 4 18These tables define a function that we will use to transform our keys into nodes in a graph. For each key,
we apply the function twice, once each for T1 and T2. Index the tables by the letter position and the letter
value, add the numbers found in the tables, and return a result modulo the modulus2
. We get two values per
key, one for T1, and one for T2.These values are taken as node IDs in our graph, and an edge runs between
them corresponding with the index of the key in the input vector, so “one” corresponds with index zero,
“two” with index one, etc.If you look carefully at the hash function defined in the project source template, you will notice that we always use tables
of size 4 × 64. We’re doing things slightly differently there, to reduce code complexity. Instead of indexing with letter position,
we index with letter position mod 4, and instead of letter value, we use letter value mod 64. The example in this document uses
a simpler (and better) approach that is easy to do when hand tuning, but difficult to automate. The fundamental structure of the
algorithm is unchanged in either case.Applying our table to the first key, “one”, and recall that we are using only the first two letters in this
example (and that we always use the entirety of every word in our program code), we have T1(0, o) ==
19 and T1(1, n) == 5. 19 + 5 ≡ 3 mod 21, so one end of our “one” edge is node 3. The other end is
given by T2(0, o) == 15 and T2(1, n) == 19, and 15 + 19 ≡ 13 mod 21. So the “one” edge runs from
node 3 to node 13.The following table gives the calculation for the entire input set:
key T1(0,letter) T1(1,letter) sum mod 21 T2(0,letter) T2(1,letter) sum mod 21
one 19 5 3 15 19 13
two 0 6 6 8 18 5
three 0 0 0 8 2 10
four 15 6 0 2 10 12
five 15 5 20 2 6 8
six 13 5 18 3 6 9
seven 13 3 16 3 16 19
eight 8 5 13 20 6 5
nine 10 5 15 11 6 17
ten 0 3 3 8 16 3The T1 sum column is also called u(key) and the T2 sum is v(key). We’ll be referring to them by these
names later in the algorithm.Now our graph is built. The next step is to check if the graph is suitable to build our hash table. A
graph is suitable if it contains no cycles. Cycle detection is usually done with a depth first search, but in this
example, we’ll inspect by eye; in program control we would call a method for this check, one that you will
be implementing!Look at the final row in the above table and note that the “ten” edge is a self loop; it both begins and
ends on node 3. Therefore, we need to discard this graph and the associated tables. We generate a new set
of random tables:
T1
e f h i n o s t w
1 13 16 3 3 7 10 1 17
19 11 17 6 20 3 0 17 14
T2
e f h i n o s t w
7 5 6 13 20 8 19 10 16
8 3 0 1 13 0 17 14 4
and calculate a new graph:key T1(0,letter) T1(1,letter) sum mod 21 T2(0,letter) T2(1,letter) sum mod 21
one 7 20 6 8 13 0
two 1 14 15 10 4 14
three 1 17 18 10 0 10
four 13 3 16 5 0 5
five 13 6 19 5 1 6
six 10 6 16 19 1 20
seven 10 19 8 19 8 6
eight 1 6 7 7 1 8
nine 3 6 9 20 1 0
ten 1 19 20 10 8 18Applying our cycle-detecting eyeballs, we see that this graph has no loops:
9
0
6
19
8
7
15
14
5
16
20
18
10
four (3)
nine (8)
one (0)
five (4)
seven (6)
eight (7)
two (1)
three (2)
six (5)
ten (9)The final step is to fill the array that defines the g() function (as named in the paper; “g” does not stand
for “graph”). g() maps edges in our graph to indices in our original input array. We do this by traversing
the connected components of the graph. We always start with the lowest-numbered, unvisited vertex. When
traversing through a node where you have a choice of which direction to travel next, the choice does not
matter.We start with node zero and assign g(0) = 0. Then we traverse the neighbors of node 0; since neighbor
order doesn’t matter, let’s do node 9 next. g(9) = 8 − g(0) mod 10 ≡ 8, where the 8 in the subtraction
comes from the edge value and 10 is the number of keys in the input vector. Going back to node 1 and
heading the other direction, we have g(6) = 0 − g(0) mod 10 ≡ 0; again, the 0 in the subtraction is the
edge value. Continuing to node 8, g(8) = 6 − g(6) mod 10 ≡ 6. Node 7, g(7) = 7 − g(8) mod 10 ≡ 1.And node 19, g(19) = 4 − g(6) mod 10 ≡ 4.
This completes the first connected subgraph. Let’s move to the next subgraph. The lowest-numbered,
unvisited node is node 5. We assign g(5) = 0, and proceed as above. Node 16, g(16) = 3 − g(5)
mod 10 ≡ 3. Node 20, g(20) = 5 − g(16) mod 10 ≡ 2. Node 18, g(18) = 9 − g(20) mod 10 ≡ 7. And
node 10 completes this subgraph, g(10) = 2 − g(18) mod 10 ≡ 5.There is one more subgraph, containing nodes 14 and 15. Start with the smaller, g(14) = 0, and
g(15) = 1 − g(14) mod 10 ≡ 1.
And here is g:
index 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
g 0 0 0 0 0 0 0 1 6 8 5 0 0 0 0 1 3 0 7 4 2Positions in this array that were not assigned in the last step do not matter. We choose to fill them in with
zeros.As stated above, u(key) is the first “sum mod 21” in the final graph calculation table above, and v(key)
is the second such column. Our hash function is given by:
HASH(key) = (g (u (key)) + g (v (key))) mod 10
thus:
HASH(“seven”) = g(8) + g(6) mod 10
= 6 + 0 mod 10
= 6
which is the seventh element, since arrays are zero indexed; and
HASH(“three”) = g(18) + g(10) mod 10
= 7 + 5 mod 10
= 12 mod 10
= 2A Second Example
Here we present one more example, this one from the paper. In this example, the keys are the English
month names, with sub-keys being the second and third letters, as these are sufficient to unambiguously
differentiate between the names. Again, remember that the concept of a “sub-key” is for illustration only!The Java implementation has no such notion.
Key Minimum Unique Sub-key
January an
February eb
March ar
April pr
May ay
June un
July ul
August ug
September ep
October ct
November ov
December ecThe modulus is chosen to be 25, again, one more than twice the number of keys.
We populate our tables, T1 and T2, with random numbers less than the modulus. This example omits
values that will never be used to emphasize that those values don’t matter. For instance, note that none of
our sub-keys starts with a b or ends with an o:
T1
a b c e g l n o p r t u v y
11 1 13 12 17 1
9 21 13 5 19 20 1 0 3 12
T2
a b c e g l n o p r t u v y
5 2 21 24 8 12
9 23 5 2 7 12 17 2 11 8Applying the tables to our keys:
key T1(0,letter) T1(1,letter) sum mod 25 T2(0,letter) T2(1,letter) sum mod 25
January 11 19 5 5 7 12
February 13 9 22 12 9 5
March 11 1 12 5 17 22
April 17 1 18 8 17 0
May 11 12 23 5 8 13
June 1 19 20 12 7 19
July 1 5 6 12 2 14
August 1 13 14 12 5 17
September 13 20 8 21 12 8
October 1 0 1 21 2 23
November 12 3 15 24 11 10
December 13 21 9 21 23 19September has a self loop, but there is a also a cycle between January, February, and March on nodes 5,
12, and 22, so this graph is unsuitable and the tables must be regenerated:
T1
a b c e g l n o p r t u v y
19 3 14 7 20 24
11 21 15 14 10 3 2 17 1 15
T2
a b c e g l n o p r t u v y
3 13 7 11 21 22
10 12 19 3 10 2 8 1 24 15
and a new graph must be calculated:
key T1(0,letter) T1(1,letter) sum mod 25 T2(0,letter) T2(1,letter) sum mod 25
January 19 10 4 3 10 13
February 14 11 0 7 10 17
March 19 2 21 3 8 11
April 20 2 22 21 8 4
May 19 15 9 3 15 18
June 24 10 9 22 10 7
July 24 14 13 22 3 0
August 24 15 14 22 19 16
September 14 3 17 7 2 9
October 3 17 20 13 1 14
November 7 1 8 11 24 10
December 14 21 10 7 12 19Here is its graphical representation, cycle free:
21 11
17
10
8 19
0
4
22
13
20
14
16
18
7
9
March (2)
January (0)
February (1) September (8)
July (6)
April (3)
November (10)
December (11)
June (5)
May (4)
August (7)
October (9)And the last step is to generate g(). As above, we start with node zero and assign g(0) = 0. Then
we traverse the neighbors of node 0. We have twelve keys in this set, so we use mod 12, not mod 10 like
we did in the first example. We’ll start on the right, g(17) = 1 − g(0) mod 12 ≡ 1. g(9) = 8 − g(17)
mod 12 ≡ 7. Going up, g(18) = 4 − g(9) mod 12 ≡ 9. And down, g(7) = 5 − g(9) mod 12 ≡ 10.
Now going left from node 0, g(13) = 6 − g(0) mod 12 ≡ 6. g(4) = 0 − g(13) mod 12 ≡ 6. And
g(22) = 3 − g(4) mod 12 ≡ 9.There remain three more connected subgraphs. We leave these as an exercise. Remember to start at the
smallest node value in each subgraph and assign it zero.
Here is the final g, with the unused values omitted:
index 0 4 7 8 9 10 11 13 14 16 17 18 19 20 21 22
g 0 6 10 0 7 10 0 6 0 7 1 9 1 9 2 9Once again, here’s our hash function (the only thing different from above is the modulus, which is the
number of keys):
HASH(key) = (g (u (key)) + g (v (key))) mod 12
Let’s hash some keys (Remember that we are using the second and third characters in this example!):
HASH(“February”) = g(0) + g(17) mod 12
= 0 + 1 mod 12
= 1
and
HASH(“November”) = g(8) + g(10) mod 12
= 0 + 10 mod 12
= 10Requirements
The code template implements most of the “hairier” parts of this algorithm; indeed, you could complete this
assignment without ever putting in the time or effort to understand the algorithm, and that’s probably the
optimal approach to take with respect to your total time, but it’s sub-optimal with respect to your learning.You will be implementing most of the graph functionality, including cycle detection, and some code generation (much of code generation is done for you, partly so that you have example code to look at, and partly
so that you have working functions in your generated output that you may use to test it.Required classes, interfaces, and methods
All classes, interfaces, and methods appearing in the template are required and described therein.
Edge.java, Vertex.java, and Visitable.java all describe interfaces, and you will not be modifying
these files (though you will be implementing methods of these interfaces).Input
main() takes one required and two optional arguments. The required argument contains the keys, one per
line. Note that if the file contains any repeated keys, the resultant graph will always have a cycle, so your
program will go into an infinite loop.The optional arguments to main() are a prefix and a seed. By default, the program will generate a
class named CHM92Hash in a file named CHM92Hash.java. This makes in impossible to use multiple of
these hash tables in the same program (without manually editing the files). The prefix makes the manual
edits unnecessary; a prefix of foo results in the class fooCHM92Hash and file fooCHM92Hash.java. The
seed parameter allows you to get reproducible behavior from the random number generator by forcing a
particular seed. This is useful for debugging. You cannot specify a seed without using a prefix.Classes
Templates for the following classes are provided and will require modification where noted in the source.PerfectHashGenerator
This class contains the main() method. The class is fully implemented and should not be modified. The two
generate() methods build graphs and run the code generator. readWordFile() reads the input. mapping()
handles all of the algorithm steps through cycle testing.Graph
This class implements the graph data structure. A toString() method is implemented for you so that you
can easily print and inspect your graphs while you debug. You will be implementing methods to add edges
to the graph, to mark nodes as visited or unvisited, and to test for cycles.Specific to the algorithm in this class is the fillGArray() method, which traverses the final graph and
calculates the g() function, as described above. You will not have to modify this method.CodeGenerator
Given T1, T2, g(), and the modulus, we can write a Java file that implements our hash table. The code
generator also takes the key list in order to write it to the output source file, but this isn’t strictly necessary
for the hash table to function (see the comments in the output). You will complete the implementation of
the begin(), array(), table(), and end() methods, all of which write code to the output file.Testing your code
Once you are successfully writing code that compiles and that you believe to be correct, uncomment the
generated main() method, compile, and run. It should produce no output. If it produces output, there is an
error in your code.Once you believe your solution is working, modify one or more of the keys in a generated KEY LIST
array, recompile, and run again. Your code should (probably3
) print output about hashing errors, one for
each key changed.I can run the solution on /usr/dict/words (a word list that most UNIX and UNIX-like systems have)
on my workstation, which contains 38619 keys. This takes less than 2 seconds to complete. When I then
attempt to compile the output, I get:It’s possible (roughly 1 in number of keys probability) that your change will result in a new key that maps to the same index, in
which case you won’t see an error message.
sheaffer@sheaffer-pc:˜/228-f2020-assignment5/src$ javac CHM92Hash.java
CHM92Hash.java:8: error: code too large
public static final String[] KEY_LIST = {
ˆ
CHM92Hash.java:1: error: too many constants
public class CHM92Hash {
ˆ
2 errors
sheaffer@sheaffer-pc:˜/228-f2020-assignment5/src$This is the result of a hard limit that is coded into the JVM. There is no workaround; Google the error
message for details. Some testing shows that the maximum input size in Java to allow compilation of the
output is about 2500 words.I wrote a C version of this algorithm which I’ve used to create, compile, and run perfect hash tables that
contain over 4 million keys.Submission
You are required to include, in your submission, the source code for each of the classes and interfaces in the
code template, as well as any additional classes or methods you may have written to complete the assignment
(you shouldn’t need any).You need to write proper documentation with JavaDoc for each method in each
class. Write your class so that its package name is edu.iastate.cs228.hw5. Your source files (.java files)
will be placed in the directory edu/iastate/cs228/hw5 (Linux) or eduiastatecs228hw5 (Windows),
as defined in the template code. Be sure to put down your name after the @author tag in each class source
file. Your zip file should be named Firstname_Lastname_HW5.zip.